In the circuit shown in figure C = 6 µF. The charge stored in capacitor of capacity C is

zero 90 µC 40 µC 60 µC

40 µC

Both the capacitors are in series. Therefore charge stored on them will be same. Net capacity = $\frac{(C)(2 C)}{C+2 C}=\frac{2}{3} C=\frac{2}{3} \times 6 \mu F$ = 4 µF Potential difference = 10V ∴ q = CV = 40 µC